By Siemons J.
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Additional info for 2-Designs and a differential equation
The deﬁnition certainly implies this, but is considerably stronger. For example, consider the sequence an = (−1)n . Then, given any ε > 0, there is an N ∈ Z+ such that |an − 1| < ε. For example, N = 2 will do for any ε > 0, since |a2 − 1| = |1 − 1| = 0 < ε. But clearly an does not converge to 1. In fact, an does not converge to anything. Can we prove this? 8. Claim: the sequence an = (−1)n does not converge (to any limit). Proof. Assume, to the contrary, that an → L. Then, given any positive number ε, there exists N ∈ Z+ such that |an − L| < ε for all n ≥ N .
As far as we’re concerned, R is, by deﬁnition, a complete ordered ﬁeld (so we are not deﬁning it to be the set of all decimal expansions, for example). ” Well, to be a ﬁeld, it must contain a distinguished element 1R whose deﬁning property is that 1R × x = x for all x ∈ R. We can identify this special element with 1 ∈ Z+ . , and, by the properties of the ordering relation, these are all distinct elements. Arguing similarly, one may show that R must contain Z and Q. 18 (The Archimedean Property of R).
7), so Q is countable. page 15 September 25, 2015 16 17:6 BC: P1032 B – A Sequential Introduction to Real Analysis A Sequential Introduction to Real Analysis We next prove that the set R is uncountable. It’s not hard to show that the set of all decimal expansions is uncountable. But, as far as we’re concerned, R is just a complete ordered ﬁeld and there is no obvious reason why every element of such a ﬁeld should be representable by a decimal expansion. To prove that R is uncountable without the crutch of decimal expansions, we need to introduce the idea of nested intervals.
2-Designs and a differential equation by Siemons J.